Question

For fully developed laminar pipe flow in a circular pipe, the velocity profile is u(r) = 2(1-r2 /R2 ) in m/s, where R is the inner radius of the pipe. Assuming the pipe diameter is 4 cm, find the maximum and average velocities in the pipe as well as the volume flow rate

Answer 1

Answer:

a) $v_{max} = 2\ \textup{m/s}$

b) $v_{avg} = 1\ \textup{m/s}$

c) Q = 1.256 × 10⁻³ m³/s

Explanation:

Given:

The velocity profile as:

$u(r) = 2(1-\frac{r^2}{R^2} )$

Now, the maximum velocity of the flow is obtained at the center of the pipe

i.e r = 0

thus,

$v_{max}=u(0) = 2(1-\frac{0^2}{R^2} )$

or

$v_{max} = 2\ \textup{m/s}$

Now,

$v_{avg} = \frac{v_{max}}{2}\ \textup{m/s}$

or

$v_{avg} = \frac{2}}{2}\ \textup{m/s}$

or

$v_{avg} = 1\ \textup{m/s}$

Now, the flow rate is given as:

Q = Area of cross-section of pipe × $v_{avg}$

or

Q = $\frac{\pi D^2}{4}\times v_{avg}$

or

Q = $\frac{\pi 0.04^2}{4}\times 1$

or

Q = 1.256 × 10⁻³ m³/s